A gas dissolves in $500 g$ of water to give a solution containing $0.0150 mol$ of the gas. For this gas $K_{H} = 3000 bar$ with $p = K_{H} x$ ( $x$ is the mole fraction in solution). Taking $M (H X_{2} O) = 18 g mo l^{- 1}$ , the equilibrium partial pressure of the gas above the solution is

Answer: C

Moles of water

= 500/18 = 27.78

x = 0.0150/ (0.0150 + 27.78) = 5.397 \times 1 0^{- 4}

p = K_{H} x = 3000 \times 5.397 \times 1 0^{- 4} = 1.62 bar

. Using a 1 kg basis for the water would halve this value.